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The square root of any positive integer is either an integer or an irrational number. So, \(\sqrt{integer}\) cannot be a fraction, for example it cannot equal to 1/2, 3/7, 19/2, ... It MUST be an integer (0, 1, 2, 3, ...) or irrational number (for example \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{17}\), ...). So, as given that \(x\) is a positive integer then \(\sqrt{x}\) is either an integer itself or an irrational number.

(1) \(\sqrt{36x}\) is an integer:

\(\sqrt{36x} = 6\sqrt{x}\). As discussed above, \(\sqrt{x}\) cannot be a fraction, say 1/6 and integer*irrational cannot be an integer. Therefore, for this statement to be true, \(\sqrt{x}\) must be an integer. Sufficient.

(2) \(\sqrt{3x+4}\) is an integer. If x = 4, then the answer is YES but if x = 7, then the answer is NO. Not sufficient.